Using the intersheaths, maintaining at different potential, uniform distribution of stress is obtained in the cables.
Consider a cable with core diameter d and overall diameter with lead
sheath as D. Let two intersheaths are used having diameter d1 and d2 which are kept at the potentials V1 and V2 respectively.
The intersheaths and stress distribution is shown in the Fig. 1.
Fig. 1 |
Let V1 = Voltage of intersheath 1 with respect to earth
V2 = Voltage of intersheath 2 with respect to earth
It has been proved that stress at a point which is at a distance x is inversely proportional to distance x and given by,
Where k is constant.
So electric stress between he conductor and intersheath 1 is,
Now potential difference between core and the first intersheath is V-V1.
Substituting in equation (2) we get,
Now this stress is maximum at x = d/2, on core surface.
Similarly potential difference between intersheath 1 and intersheath 2 is V1 - V2.
Now g1 will be maximum at the surface of intersheath 1 i.e. x = d1/2.
The potential difference between intersheath 2 and outermost sheath V2 is only as potential of intersheath is maintained at V2 with respect to earth.
This g3 will be maximum at x = d2/2
Choosing proper values of V1 and V2, g1max, g2max etc. can be made equal and hence uniform distribution of stress can be obtained.
The stress can be made to vary between same maximum and minimum values as shown in the Fig. 1, by choosing d1 and d2 such that,
d1/d = d2/d1 = D/d2 = α
and g1max = g2max = g3max
Let us try the express voltages V1 and V2 interms of V and α.
Substituting value of V2 from equation (11),
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